一、内容
Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.
* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.
If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?
Input
Line 1: Two space-separated integers: N and K
Output
Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.
Sample Input
5 17
Sample Output
4
二、思路
- 用vis数组记录当前点是否被访问过，被访问就不用再入队了。
- 每次3种状态 *2 +1 -1
- 若n大于k的话只能向后移动 输出n-k即可
三、代码
#include <cstdio>
#include <queue>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn = 1e5 + 5;
int n, k, step[maxn], vis[maxn];
void bfs() {
queue<int> q;
q.push(n);
vis[n] = 1;
while (q.size()) {
int t = q.front();
q.pop();
int next;
for (int i = 0; i < 3; i++) {
if (i == 0) next = t * 2;
else if(i == 1) next = t + 1;
else next = t - 1;
if (next <= 0 || next >= maxn) continue;
if (!vis[next]) {
q.push(next);
vis[next] = 1;
step[next] = step[t] + 1;
}
if (next == k) {
printf("%d", step[k]);
return;
}
}
}
}
int main() {
scanf("%d%d", &n, &k);
if (n >= k ){
printf("%d", n - k);
} else {
bfs();
}
return 0;
}
